Answer
$403\space rev/day$
Work Step by Step
Let's assume this star is a solid sphere. In this case, there's no external torque. So its angular momentum is conserved.
We can write,
$I_{1}\omega_{1}=I_{2}\omega_{2}-(1)$
For a solid sphere, Rotational Inertia $(I)=\frac{2}{5}MR^{2}-(2)$
$(2)=\gt(1)$
$\frac{2}{5}MR_{1}^{2}\omega_{1}=\frac{2}{5}MR_{2}^{2}\omega_{2}$
$\omega_{2}=\omega_{1}(\frac{R_{1}}{R_{2}})^{2}$ ; Let's plug known values into this equation.
$\omega_{2}=\frac{1\space rev}{34.4\space day}(\frac{4.96\times10^{8}m}{4.21\times10^{6}m})^{2}= 403\space rev/day$
